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 05-29-09, 10:29 PM #1 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Rudimentary question about finding upper/lower bound of integral Okay I am taking Calculus 2 this summer at my local university. It's a summer course so it's faster and therefore harder. I don't have a lot of access to help Use the min-max inequality to find the upper and lower bounds for the value of the integral from 0 to 1 of 1/(1-t^2) the lower bound can be found by f(minimum) * (b-a) and the upper by f(maximum) * (b-a) So when they say f(min) and max, what exactly do they mean? To find a function's minimum I would set the first derivative equal to zero. Do I plug this x value -- where the minimum occurs at -- or the y value, what the minimum value is, into the equation? __________________ C://dos C://dos.run run.dos.run
 05-29-09, 11:10 PM #2 MrCrassic  Senior Member     Join Date: Jun 2007 Location: Brooklyn, NY Bikes: 2008 Giant OCR1 (with panda bear on the back!) Posts: 3,650 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) I believe you input the minimum value that you found. I might be wrong; someone please verify. __________________ Ride more. Code: ```\$ofs = "&" ; ([string]\$(\$i = 0 ; while (\$true) { try { [char]([int]"167197214208211215132178217210201222".substring(\$i,3) - 100) ; \$i = \$i+3 > catch { break >>)).replace('&','') ; \$ofs=" " # Replace right angles with right curly braces```
 05-29-09, 11:52 PM #3 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) So by this you mean what the minimum/maximum is, the y value? I just want to be 100% clear. I think you're right. __________________ C://dos C://dos.run run.dos.run
 05-30-09, 08:42 AM #4 apricissimus  L T X B O M P F A N S R     Join Date: Jun 2008 Location: Malden, MA Bikes: Bianchi Volpe, Bianchi San Jose, Redline 925 Posts: 2,334 Mentioned: 2 Post(s) Tagged: 0 Thread(s) Quoted: 1638 Post(s) No, this is wrong! Plug in the x-value where the min/max occurs at. Suppose f(x) has a maximum on [a,b] at x_1 and a minimum at x_2. Then the upper bound your looking for would be f(x_1)*(b-a) and the lower bound would be f(x_2)*(b-a). This is equivalent to max*(b-a) and min*(b-a), respectively. Note that all you're doing is finding the smallest rectangle that entirely covers the actual area of the integral for the upper bound, and the largest rectangle that is entirely covered by the actual area of the integral for the lower bound. Last edited by apricissimus; 05-30-09 at 08:45 AM.